Question

An aluminum tensile test specimen has a diameter, d_o = 25 mm and a gauge length of L_o = 250 mm. If a force of 175 kN elongates the gauge length by 1.25 mm, the modulus of elasticity of the material is nearly
1. 71 GPa
2. 71 MPa
3. 142 GPa
4. 142 MPa

Answer 1

Correct Answer - Option 1 : 71 GPa

Concept:

The deformation of a bar due to point load is given as,

\(δ l=\frac{PL}{AE}\)

where, δl = elongation of a bar, P = load or force, L = length of a bar, A = cross-section of a bar, E = modulus of elasticity

cross-section of a bar, \(A=\frac{\pi}{4}D^2\)

Calculation:

Given:

δl = 1.25 mm, P = 175 kN = 175 × 10³ N, L = 250 mm, \(A=\frac{\pi}{4}25^2=490.87~mm^2\)

⇒ \(1.25=\frac{175~×~10^3~×~250}{490.87~\times~E}\)

Therefore, E = 71301.97 MPa = 71.30 GPa